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Question -

.┬аFind the sum of the following series:
(i) 2 + 5 + 8 + тАж + 182

(ii) 101 + 99 + 97 + тАж + 47

(iii) (a тАУ b)2┬а+ (a2┬а+ b2) +(a + b)2┬а+ sтАж. + [(a + b)2┬а+ 6ab]



Answer -

(i)┬а2 + 5 + 8 + тАж + 182

First term, a = a1┬а=2

Common difference, d =a2┬атАУ a1┬а= 5 тАУ 2 = 3

an┬аtermof given AP is 182

an┬а= a+ (n-1) d

182 = 2 + (n-1) 3

182 = 2 + 3n тАУ 3

182 = 3n тАУ 1

3n = 182 + 1

n = 183/3

= 61

Now,

By using the formula,

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

тИ┤ The sum of the seriesis 5612

(ii)┬а101 + 99 + 97 + тАж + 47

First term, a = a1┬а=101

Common difference, d =a2┬атАУ a1┬а= 99 тАУ 101 = -2

an┬аtermof given AP is 47

an┬а= a+ (n-1) d

47 = 101 + (n-1)(-2)

47 = 101 тАУ 2n + 2

2n = 103 тАУ 47

2n = 56

n = 56/2 = 28

Then,

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

тИ┤ The sum of the seriesis 2072

(iii)┬а(a тАУ b)2┬а+(a2┬а+ b2) + (a + b)2┬а+ sтАж. + [(a +b)2┬а+ 6ab]

First term, a = a1┬а=(a-b)2

Common difference, d =a2┬атАУ a1┬а= (a2┬а+ b2)тАУ (a тАУ b)2┬а= 2ab

an┬аtermof given AP is [(a + b)2┬а+ 6ab]

an┬а= a+ (n-1) d

[(a +b)2┬а+ 6ab] = (a-b)2┬а+ (n-1)2ab

a2┬а+ b2┬а+2ab + 6ab = a2┬а+ b2┬атАУ 2ab + 2abn тАУ 2ab

a2┬а+ b2┬а+8ab тАУ a2┬атАУ b2┬а+ 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Then,

S = n/2 (a + l)

= 6/2 ((a-b)2┬а+[(a + b)2┬а+ 6ab])

= 3 (a2┬а+b2┬атАУ 2ab + a2┬а+ b2┬а+ 2ab + 6ab)

= 3 (2a2┬а+2b2┬а+ 6ab)

= 3 ├Ч 2 (a2┬а+b2┬а+ 3ab)

= 6 (a2┬а+b2┬а+ 3ab)

тИ┤ The sum of the seriesis 6 (a2┬а+ b2┬а+ 3ab)

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